Solve 4/4-9i

$$\frac { 4 } { 4 - 9 i }$$

Evaluate

$$\frac{16}{97}+\frac{36}{97}i\approx 0.164948454+0.371134021i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 4+9i.

$$\frac{4\left(4+9i\right)}{\left(4-9i\right)\left(4+9i\right)}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{4\left(4+9i\right)}{4^{2}-9^{2}i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{4\left(4+9i\right)}{97}$$

Multiply 4 times 4+9i.

$$\frac{4\times 4+4\times \left(9i\right)}{97}$$

Do the multiplications in 4\times 4+4\times \left(9i\right).

$$\frac{16+36i}{97}$$

Divide 16+36i by 97 to get \frac{16}{97}+\frac{36}{97}i.

$$\frac{16}{97}+\frac{36}{97}i$$

Real Part

$$\frac{16}{97} = 0.16494845360824742$$

Solution Steps

Multiply both numerator and denominator of \frac{4}{4-9i} by the complex conjugate of the denominator, 4+9i.

$$Re(\frac{4\left(4+9i\right)}{\left(4-9i\right)\left(4+9i\right)})$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(\frac{4\left(4+9i\right)}{4^{2}-9^{2}i^{2}})$$

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{4\left(4+9i\right)}{97})$$

Multiply 4 times 4+9i.

$$Re(\frac{4\times 4+4\times \left(9i\right)}{97})$$

Do the multiplications in 4\times 4+4\times \left(9i\right).

$$Re(\frac{16+36i}{97})$$

Divide 16+36i by 97 to get \frac{16}{97}+\frac{36}{97}i.

$$Re(\frac{16}{97}+\frac{36}{97}i)$$

The real part of \frac{16}{97}+\frac{36}{97}i is \frac{16}{97}.

$$\frac{16}{97}$$